text:
EFATO after V2 during initial climb AP already on
I am seeing procedures for EFATO after V1 which means that Take off with single engine and handling situation before AP is on. What if engine failure happens just after takeoff, between take off and initial climb with AP already on, flaps cleaned to zero but still climbing at green dot speed below 10000 ft?

There are procedures for each of those situations, what is your question? V1 is the maximum speed at which takeoff can be aborted. Vr is the speed at which the pilot actively starts the liftoff process and V2 is the speed at which a safe climb-out can be maintained with one engine inoperative.

Hi, unsure what is unclear. I'd like to know the procedure for an engine failure happening after v2. in initial climb stage towards FL 10 000Ft (so we are below 10000) AP already  on.also specified that already accelerating i.e after thrust red/ acc and speed towards s speed. Ideally procedure if no engine damage and the with engine damage. @bianfable any help or documentation link?

The autopilot will handle engine failures. If ATHR is on, put the thrust levers into MCT gate. Once the flight path is stabilised apply the engine failure procedure as described by the ECAM.


text:
What are these pipelines for between the folding wing mechanism of F4 Phantom II FGR2 and Buccaneer?
The top picture is from Buccaneer and rest are from F4 FGR2. There's not any sealing mechanism between the joints, so these can’t be fuel or hydraulic related. They might have something to do with the blown air slaps/flaps systems, I'm not sure.

From Dowgwillo on boundary layer control, cited elsewhere:

Yes they are engine bleed ducts, either for anti-icing purposes, or for blowing along the flap leading edge to improve the efficiency of the flap.
High temp ducts often depend on just a smooth metal flange that mates with its connecting duct to provide an adequate seal, with possibly some compression preload provided by the springyness of the duct when they are butted together. They don't have to be 100% air tight, just tight enough.

Thanks for confirming my assumption, did some further research on bleed air application on F4, turned out it did have such system. The diagram also shows 3 sub  pipelines around the folding mechanism, from main on the leading edge.

The Phantom has blown flaps?

according to this https://arc.aiaa.org/doi/pdf/10.2514/6.2018-0561

According to the following cutaway of the Buccaneer, 53 and 67 seem to be good candidates for your answer i.e. ducts for the leading edge, aileron and flaps blowing system.

According to this other cutaway, the only system located in the outer wings of the F-4 is 78 and 123 i.e. the slat actuators.

Did some further research, on F4 the bleed air system was implemented for boundary layer control,  on lead slats and flaps

@DanielYang: Very interesting discovery  So the F-4 did have a boundary layer control like the Buccaneer

Manage to find this document, unfortunately don't have the full access.
https://arc.aiaa.org/doi/pdf/10.2514/6.2018-0561

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Tail Number of a particular plane
How to determine the aircraft tail number on a particular flight, using the flight number. I have been advised there was a bird strike on a plane I was supposed to be on, but air stewards told a different story to why we were delayed.

Hello Jackie, welcome to aviation.stackexchange.com!  Can you provide more information about the flight in your question, such as flight number, date of flight, airport of departure and destination, aircraft type, airline. All that information may help users here to find the aircraft registration for you.

You can also try https://wildlife.faa.gov/search which collects voluntary strike reports. Use "export to Excel" to see the registration, the phase of flight and the damages. US only and about one report per day. Not much.

Hi thanks, the flight number was FR656, the date of flight was 6th April 23', the airport of departure was Malaga Spain and its destination was Glasgow Prestwick, Scotland UK. Not sure what type of plane it was, but the airline was Ryanair.

You can try to look the flight up on a flight tracking website such as flightradar24 or flightaware. They have information about the registration of the aircraft.
For access to historic information or aircraft registration details you may have to pay for a subscription, especially if the flight was more than a week ago.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
How is the geometry of slotted flaps and slats designed?
I am currently doing a school research project attempting to find the effect of single-slotted flaps and slats on an airfoil's (NACA 2412 or similar) lift coefficient. I will be using a small homemade wind tunnel to see the lift force that a small airfoil model generates with and without the use of high-lift devices, and backing the claims up using CFD data. The model will be a simple wing that is essentially an elongation of a 2D airfoil.
When designing the slotted flap, there seems to be very little information on how the geometry of the flaps is created nor is there much data on existing designs. There are many research papers investigating the effect of high-lift devices, but almost none explain how their designs were produced or why they used the flap that they did. Essentially, how the flap takes its shape. For example, in Theory of Wing Sections, various slotted flaps were tested for their max lift coefficients, but the book does not specify how these designs are created (see image). The same goes for slats.

My question is, how are these various designs of flaps and slats produced mathematically and how can I replicate this design process for my airfoil? The flaps and slats do not need to be perfect, but I will have to justify what led me to the design that I will be using in my experiment.
Alternatively, I can also use proven designs of high-lift devices from existing airplanes such as a C172 or experimentally optimised ones. Unfortunately, I was not able to find any of those. If anyone is aware of existing single-slotted flap and slats schematics, it would also be very much appreciated.
I am aware that someone asked a similar question but it was deemed too broad. Hope this isn't.

High lift flow and high lift design is really complex.  If you choose to become an expert aerodynamicist, you could spend your entire career working on high lift flow.
The most important paper is "High Lift Aerodynamics" by AMO Smith.  It is 30 pages long and I had to read it many many times before I started feeling like I understood many of the ideas.  That was half a lifetime ago -- if I was to need to get into high lift aero again, I would start there.
I have access to wind tunnel data for several high lift tests.  These tests were performed to be used to test, calibrate, and improve CFD codes.  The documentation I have provides all of the information you would need to either build a similar test section for your wind tunnel or to conduct your own analysis of the foils.  If you want, we can figure out how to share this data with you.
There are several high lift airfoils you could choose from.
A2, a Farnborough High Lift airfoil.  This one comes in three configurations, but all include a slat as well as flaps.

A5, a McDonnell Douglas aileron test originally conducted at relatively high Mach number (around 0.7).

A9, an NLR airfoil

Boeing TR-1332 airfoil

I would probably recommend A9, the NRL airfoil.  It avoids the complexity of the multi-element systems (including flaps).  Similarly, it avoids any transonic flow effects that the aileron test case is designed to go after.

Thank you so much for this information! I would love to have a look at the data that you mentioned if possible.

I have just downloaded these, really appreciate your help. I still have a couple of follow-up questions regarding the airfoil that you recommended. Would you be so kind to answer them?

Happy to help out.

Here we go: 1. Since this airfoil has a max thickness of 16.5%, will I be able to apply thin airfoil theory to it for lift calculations? (or which other theory should I use to calculate a theoretical lift value?) 2. If I 'borrow' the NLR airfoil design, do you advise me to ditch the slats in the investigation to simplify things? 3. To change the angle of the flap, can I simply rotate the flap at that point or do I have to do additional translation to it? 4. I understand that this is a supercritical airfoil, how different are its properties to something like a 2412 at really low airspeed?

Just for clarification as I forgot to mention in my original post, my investigation is about how the angle of flap/slat deployment changes lift coefficient (hence question 3). I do not have access to research-grade facilities, so I will just be using what looks like PC fans as a source of air flow (hence question 4). I have proven with a really simple setup that a simple 2412 model actually does generate some lift, so I was wondering how much it could change with the NLR.

Thin airfoil theory would probably work ok-ish on the single-element version of this foil (despite its thickness), but TAF requires the camber line and thickness form to be broken apart.  That is a challenge itself.  The NLR airfoil does not have a slat.  Were you planning on including a slat in your investigation?  If so, then choose the case that best meets your needs.  The NLR is not a supercritical foil -- it was designed to be a very simple test case at low speed.  The documentation of all of these cases should make clear how the deflection angle is defined.

For the NLR, there appears to be a + in front of the flap indicating the center of rotation.  Spend some time digging into the documentation I sent -- I know there is a ton of it, but read the PDF files on each case of interest and see what you think.

Yes, I was planning on using a slat. The first A2 case looked appealing to me at first. Since you recommended the A9, how does it exactly avoid the complexity of the multi-element systems?

The A9 has two total elements, but only a single element flap.  The more elements, the more complexity in general.  The A9's cove is smoothed over to not have a pocket of separated flow.  The others have sharp edges on the lower surface that will force separation pockets either in the main airfoil cove or even behind the slat.  The most simple computational methods will not capture that.

That makes sense, thank you!


text:
Pilot Jobs, where to find one? How to land one?
I have just finished doing my commercial license in US and hold an FAA multi single IR and are converting to my Australian license, what are the job prospects like for a low time pilot in Australia i am not sure weather to go back to the US and train to get a CFI job over there. Just a little lost with my aviation career at the moment and in need of some guidance.
PS I am really not fussed who or where I fly, I just want to fly and get paid for it.

Get time.  Get time any way you can in anything eith wings.  Fly pipeline patrol, fly jumpers, fly for scheduled pt135 (e. Botique Air).  CFI work is a great way to rack up hours but I'd invest in a CFII and MEI if you take that route.

There are a reasonable amount of low-hour jobs in Australia, but for every one of those jobs there are dozens of pilots. I don't know how that compares to the US but I'm sure it's similar. Very few of those jobs are in the cities. Most people get their first job by driving out to the operators in the middle of nowhere, to prove their dedication, and stay there for a few weeks until they get an opportunity. Find something that sets you apart from the rest and sell that to anyone who will listen. Make friends with people with and without jobs, because you never know what might come up.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What is this plane seen in the Falcon Crest TV series?

Does anyone recognize the make and model of this small plane?  It was used in episode 154 <6.27> "Chain Reaction" of the 1980's TV show "Falcon Crest".  I'm working on a database for the show's picture planes and I am missing the identification of this aircraft and two more planes. Thank you so much for your help.

Obviously a Cessna.  Exact model (172, 175, 177, 182) depends on details I'm not familiar enough with to call.

Looks like a Cessna 210 to me.

If the episode ever shows the tail number, you can google it and it will take you to the FAA records for the airframe showing you manufacturer, model, owner, etc. ie N12345

Thank you so much for your help.

@mikem This suggestion works for day-to-day photos, but may not always work for movies. Movies may use fake or "recycled" registrations.

That's a Cessna 210 Centurion II
.
The Cessna 210 models K, L, M, N and R have the windows shaped like the one shown in the picture from Falcon Crest. If it has a two-bladed propeller, it is a model K, the others have three-bladed props.
The picture below is a Cessna 210M.

source: www.airhistory.net, photo copyright: Phil Vabre

Thanks a lot. Your explanation is very helpful.


text:
What is this helicopter seen in the Falcon Crest TV series?

My next to last "Falcon Crest" related question for the database I'm programming: Does anyone recognize the make and model of this helicopter from episodes # 160 <7.05> (first screen capture) and # 163 <7.08> (second screen grab).
My guess is a Bell Jet Ranger, but I'm not sure.
Any hints are appreciated!  Thanks.

Yes it's a Jet Ranger.

Looks to me like a Bell 206 JetRanger/LongRanger


text:
From which aircraft are these parts (plane crash on "Dallas" TV series)?

I'm trying to identify the aircraft in this scene from episode # 014 <2.09> entitled "Survival" of the classic "Dallas" TV series (filmed and broadcast in 1978), which featured the debris after a plane crash.
The footage was reused in 1984 as "a day for night" shot for another plane crash in a "Falcon Crest" episode, so this is the last aircraft missing in the "Falcon Crest" database I'm programming.
Please note:
I am not sure if the production company used parts of a real aircraft to stage the scene or if this is a complete mock-up.  The "SF" logo (for "Southfork Ranch" on "Dallas") is definitely a prop the movie studio’s art department added to the debris.
I have no idea if it is possible to get any clues from the debris about the make and model these parts might have come from  --  if these were real airplane parts at all.
Any input is appreciated.  Thank you so much for taking a look!

I'd be surprised if there was anything in that picture that could positively identify the aircraft.

I agree with Ron.  Is your purpose in documenting the actual aircraft flown in the scenes filmed, or what they should be per the plotline?  I ask because TV and Movies are well known for mixing and matching the wrong things.  i.e. Actors filmed climbing the stairs into one kind of airplane, but different one is later shown supposedly taking off with them inside.  Would watching earlier episodes of Dallas clue you in on what type airplane JR owned?

The fin and rudder with the external ribs is pretty distinctive.  There should be prizes for correct answers on these sorts of questions.

@JohnK There is, you can sponsor a bounty. I'm not sure it's distinctive enough though to match to one model, might be able to narrow it down to a few choices... More likely than not it's just some corrugated sheet metal they bent up and painted for this scene.

My purpose is not to find out what it should be per the storyline.  I am documenting the real aircraft (or, in this case, the parts of a possibly real aircraft) they used for filming.
I am doing this for the "Falcon Crest" TV series, which reused this clip, so there's no need to investigate J.R.'s plane in the actual flight scenes on "Dallas" because post-production on "Falcon Crest" simply used this "Dallas" stock footage to illsutrate the crash site after a Grumman Gulfstream II went down  --  and it's pretty obvious these parts do not match a Gulfstream at all.

It also may be that the producers went to a junk yard and bought a wing, tail, and random parts from many different aircraft that don't belong together.

I guess I can see the tail of a C-130 in the image.

Roh, could you please specify this a bit more?  Thank you.

I think, based on that tail section, that it looks a lot like a Stinson 108-3.  They had large metal tails with horizontal corrugation.
The position of the rear nav light and trim tab (just above the light) allows for enough room below the light for the six rows of corrugation that I count in photos of other 108-3s.
The position of the tail wheel fits, too.  It is not too long but it is easily noticable and not buried in, or up close to, the tail.
The wings are metal with fabric covering.  108s were all metal and covered with fabric, except the tails.
The only thing that does not seem to match is the wing rib.  The Univair website images for 108 wing root ribs have a very similar outline but do not have what appear to be four holes near the rear edge.  They have a single large hole instead.  Univair may be selling improved parts, however, and the rib in the video may be a factory original -or- a field manufactured piece as part of a repair that does not match the original.

Upvoted for the good research, but I too have my doubts because of the wing rib. It doesn't match the pictures in the Univair catalogue and neither does it match pictures of deceased Stinson Voyagers.

I agree with acpilot.  The line of reasoning for the Stinson parts sounds very convincing after re-examining the footage in the episode in comparison with photos of Stinson 108-3 planes.  They probably used primarily Stinson parts, but may have added some other wing rib to the debris.


text:
What is this jet seen in the Falcon Crest TV series?

This is another jetliner from "Falcon Crest" I'm trying to identify.  It was visible briefly in episode # 194 <8.11> "True Confessions" (1989).  Thanks for your help!

I believe that it is an Israeli Aircraft Industries Westwind. This is based on the mid-set wing and the 5 windows on each side.

Image source: https://upload.wikimedia.org/wikipedia/commons/thumb/f/ff/IAI_Westwind_%282323424886%29.jpg/300px-IAI_Westwind_%282323424886%29.jpg

Thank you so much!


text:
What happens in order to make a shock when the flow first turns?
Imagine you have a really zoomed in view of a 2d ramp, which is at 20 degrees. Zoom in on the exact point the ramp turns from horizontal to 20 degrees.
When the first supersonic air molecules reach where the ramp turns, they rapidly compress into a shock. My question is, what determines what angle this shock will be? What mechanism makes the shock be at a certain angle?
After this first section of shock is made, I understand why it will continue outwards, but at the exact point where the first section of shock is first made, I don’t understand how it will choose its own angle, considering different turning angles and speeds.
For example if you look at the oblique shock chart, there are different angles the shock will be with different speeds/turning angles. This question is asking how those shocks will ‘choose’ their angle based on flow turning and speed, or what happens to make them at said angle.

Different way of asking:
When the first air molecules get to where the wedge turns to 20 degrees, what exactly do the air molecules do to make a 54 degree shock (at Mach 2)?
(54 degrees according to the oblique shock chart)
(This (great) answer was explaining something different; it didn’t talk about what exactly happens at the point where the flow turns)

Photos to help:

first supersonic air molecules doesn't make sense. Air molecules faster then the fastest air molecules . Tip: avoid mixing the mental model about "atomic/molecular gases", with pressure/temperature (and sonic speed): statistics is not intuitive (also for statisticians) so risk to do things wrong are high. Either you do very methodological and mathematical, or use one or the other mental models.

Doesn't the shock form at the point where the molecules go supersonic?  ("Turning" can involve acceleration)

So, first a disclaimer.  The oblique shock charts are based on inviscid flow.  This works great away from the wall, but exactly at the corner (if you zoom in like you're saying), there is of course a boundary layer.  That boundary layer messes with reality.
If we stick with inviscid flow, then the streamline at the wall is no different from the rest of the streamlines.  The inviscid wall is what we call a slip wall (a viscous wall is a no-slip wall).
The wall is just like another streamline.  It is a streamline that turns twenty degrees.  The next one out must do the same thing, etc.  If you accept that the other streamlines turn and form a shock, then the same thing applies to the wall.
In this case, the angle of the oblique shock is strictly determined by \theta and Mach number.
Post Script
If that answer isn't satisfactory (which it probably isn't), then to really get into it, you have to go through the derivation and understand all the math.
It is only conservation of mass, energy, and momentum -- with a perfect gas equation of state.  But connecting the dots from those simple principles to the end state is non trivial.
You need to read a textbook on gas dynamics or an aerodynamics textbook.  Either one should do a good job with the oblique shock derivation.  Don't just jump to that chapter, start at the beginning (of the compressible flow part if an aerodynamics book) and go through 1D flow, normal shocks, and then oblique shocks.
Edit: Trying to explain the shock angle a little more...
Here is something that blew my mind when I learned it.....
When you consider an oblique shock, you decompose the velocity vectors into two components -- the component normal to the shock and the component tangential (parallel) to the shock.  (Don't worry about the fact that we don't know what the shock angle is yet).

Here the incoming velocity is V1.  Like usual, the shock angle is $\beta$.  The component of V1 normal to the shock is $V_{n1}$ and the component of V1 tangent to the shock is $V_{t1}$.
Likewise, after the shock, the velocity is V2, with components $V_{n2}$ and $V_{t2}$.
Here is the mind blowing part...
$$V_{t2}=V_{t1}$$
The tangential velocity after the shock is the same as the tangential velocity before the shock!
Here is the next mind blowing part...
In the direction normal to the shock -- the flow behaves exactly as if it was going through a normal shock!
The normal Mach number before the shock $M_{n1}$ is greater than 1.0 (supersonic) and the normal Mach number after the shock $M_{n2}$ is subsonic.  The pressure, density, and temperature jumps all behave exactly like a normal shock as if the tangential velocity was not there.
Here is another view that may help...

If you were observing in a coordinate system attached to the shock, with the flow moving by you left to right, this is what you would see.
Next, imagine you were moving vertically at $V_{t1}$ such that $V_{t1}$ appeared to be zero to you.  You would see exactly the normal shock situation.  This is what the flow normal to the shock experiences.
So, if your guess was right (about the shock being parallel to the ramp), then $V_{n2}$ would have to be zero.  The flow can't penetrate the ramp.
This is what I meant by 'where would the flow go'?
$V_{n2}$ must be non-zero and $V_{t2}=V_{t1}$.  Together, these two components of velocity must form a triangle at the angle $\theta$.
Remember when I said "Don't worry about the fact that we don't know what the shock angle is yet".  This is where the math gets a bit tricky.
To solve this problem (figure out all the angles, etc), we end up with a system of multiple equations and multiple unknowns.
We know $M_1$ and $\theta$, but we don't know $\beta$.  Since we don't know $\beta$, we can't decompose $V_1$ into $V_{n1}$ and $V_{t1}$
We know that $V_{n2}$ and $V_{t2}$ must make the angle $\theta$, but we don't know their magnitudes.
We know that $V_{n1}$ and $V_{t1}$ must be the legs of a right triangle with magnitude $V_1$.
Then we know the normal shock relations for how $V_{n2}$ must relate to $V_{n1}$.
All this is non-trivial to solve and doesn't really lend itself to a simple physically intuitive solution.
In the days before computers, solving this was a real pain.  So, really smart people at NACA (the predecessor to NASA) did the calculations for us and produced the famous theta-beta-Mach chart.  They also produced numeric tabulated results.  These were printed in volumes like NACA 1135.  Paper copies of these were distributed widely -- to libraries at universities and research institutions as well as companies working on aerospace problems.
There are plenty of online copies available, but they're typically scans made from an old microfiche.  They'll do, but they're grainy and low resolution.  If you can lay your hands on an original paper copy (say on eBay), you'll find that the charts were quite beautiful.

This answer is definitely satisfactory, but it would 100% make sense if the shock angle was equal to the flow turning angle, which it’s not. Essentially I understand why the shock will form, but it doesn’t make as much sense for the shock to not equal the flow turning angle. (It does make sense, but I don’t know how it forms at not the flow turning angle)

@Wyatt it might make a lot more sense to consider that the object and the attached boundary layer are supersonic, and when they are moved into surrounding air, a shock wave will form.  The amount of turning may be related, but not exactly matching, to the oblique shock angle.

@Wyatt If the shock was the same as the turning angle, then where would the air go? The shock is not an impenetrable barrier.  It is not a wall, it is a rapid change in properties.


text:
What temperature is the vapor that forms around a maneuvering aircraft?
At what temperature does the vapor seen on aircraft maneuvering form? I know this highly depends on humidity, but is there a general calculation?

When doing my research, I found this page, but it only really talked about how air can't turn into water.
I'm fairly sure that the temperature this happens at is the dew point, but I'm not 100% sure.

I may be nitpicking, but technically, it is probably mist or some other word describing suspended droplets formed by condensation, while "vapor" in a scientific sense is a usually transparent gas...

I think I've linked to this resource in one of your previous questions.
NASA SP 514
On page 5 (pdf page 15), he derives an expression for the relative humidity around an aircraft.
$$ RH_l=RH_\infty\, \frac{P_l}{P_\infty}\,10^{\left\{-\frac{2263}{T_\infty}  \left[\frac{(T_l/T_\infty)-1}{T_l/T_\infty}  \right]\right\}}$$
Sorry the font gets so small in the exponent.  I'll write it another way...
$$ A={-\frac{2263}{T_\infty}  \left[\frac{(T_l/T_\infty)-1}{T_l/T_\infty}  \right]}$$
$$ RH_l=RH_\infty\, \frac{P_l}{P_\infty}\,10^A$$
The temperature $T_\infty$ must be in Kelvin for the number $2263$ to be correct.
The subscript $l$ means local -- at some point in the flow around the aircraft.  The subscript $\infty$ means freestream -- at some point well ahead of the aircraft.  I.e. the ambient conditions.
He is calculating $RH_l$, the local relative humidity.
Condensation begins when $RH_l>1$.  The line between clear air and air with visible water vapor should be at $RH_l=1$.
We can immediately see that when the condensation will be visible depends on the ambient humidity $RH_\infty$, ambient pressure $P_\infty$, and ambient temperature, $T_\infty$.  Furthermore, we can see that it also depends on the local pressure $P_l$ and temperature $T_l$.

I'm fairly sure that the temperature this happens at is the dew point, but I'm not 100% sure.

You are correct.

Is this true also for rocket launches where vapor sometimes appears as a rocket goes transonic?

@StevePemberton, unless there is some nuance I'm missing I don't see how it would not be.

@MichaelHall Yeah it should be as the vapor cloud around an aircraft going transonic is lower pressure, so I'd assume it is with rockets also.

So does that mean if you could touch it with your hands, the vapor would be cold or hot? ;)

@J.ScottElblein, it depends on the temperature.  ;)

lmao. Generally speaking though; hot or cold? And if hot, what degree burn would it give?

@J.ScottElblein, ok I thought you were pulling my leg with the winky face.  Comments don't give space for a full explanation, but the dewpoint will fall within normal atmospheric temperature ranges.  And always lower than ambient.  https://en.wikipedia.org/wiki/Dew_point


text:
Why can't you hear an aircraft traveling at Mach 0.95 until it passes?
So say an aircraft is traveling at Mach 0.95 towards you. It's almost going faster than the sound waves it's making.
Now say there's a speaker or something on the nose of the aircraft facing forward. If the aircraft is traveling at Mach 0.95, it still isn't going faster than the sound waves it's making. So that means the sound waves would still slowly pull away from the aircraft going forward, right? If that's the case, wouldn't you hear an aircraft going Mach 0.95 before it passes?
Like if you're on a highway going 60mph, and another car is going 65 and passes you slowly. Pretend the second car is the sound wave that you're emitting.
So to summarize, if the aircraft was going Mach 0.95, and the sound waves would slowly travel forward, wouldn't that make it so you could hear the aircraft before it got to you?
(In a video of a F/A-18 I saw, it was traveling at what I assume was ~Mach 0.9 speeds, because you couldn't hear any sound from it until it was really close. No shockwaves or anything so it wasn't going supersonic)

You are never hearing the noise the aircraft is making right now.  Instead, you're hearing the noise the aircraft made some time ago -- where that time is determined by the speed of sound and how far away you were.
This is very much related to the Doppler effect, but instead of focusing on the shift in observed frequency, there is also an amplitude effect.
Check out this website with a Doppler simulator.  Here is a screenshot from a case I played with.  You should really check out the animation.

I placed the source initial position at the far left (-50) and the observer at the far right (50).  I set the speed of the source at M=0.9 (it won't let you set .05 increments).
The pulses are generated at a uniform frequency -- you're supposed to observe the doppler effect, the fact that the pulses are bunched up in front and spread out behind.  This results in the frequency shift you hear in a passing race car.
However, the other effect is that at the moment pictured (where the first sound wave is just about to reach the observer), you're actually listening to a sound that was generated 100 m away.  The intensity of a sound drops off with the inverse square of the distance, so doubling the distance you cut the intensity by 4.
Your eyes trick you -- you're seeing the aircraft very close, but you're actually listening to the sound made when it was far away.
In our case, the sound was produced 10x as far away as what we are seeing.  Which means that the sound intensity will be 1/100th as intense as we would expect for a nearby aircraft.

I see, thanks. So if the waves that travel forward in front of the object  slowly, that means that those waves will slowly ‘reach out’ if that makes sense. If this is the case, why can you only hear a jet after it reaches close to you going just below supersonic? Wouldn’t those waves traveling forward or ‘reaching out’ make it so you could hear the jet earlier?

The waves that are reaching out are the ones in front of the object that slowly progress in the direction of travel of the object, as seen in the simulation you provided.

(The video of the jet passing by I saw was an F/A-18 that was going almost supersonic and you couldn’t hear it until it was almost right beside the camera)

The waves travel at the same speed no matter what.  They travel at the speed of sound.  Period.  Full stop.  If we approximate the speed of sound as 1000 ft/s and the aircraft is traveling at M 0.95, then when you hear the sound produced one second ago, you're hearing the sound produced 1000 ft. away.  Since that time, the sound traveled 1000 ft, and the airplane traveled 950 ft.  So you're seeing an airplane 50 ft. away, but hearing it 1000 ft. away -- which is why it sounds quiet.  The airplane is still making noise (when it is 50 ft away), but that sound hasn't reached you yet.

Ah okay, that makes sense. As you said, your eyes definitely trick you. Thanks.

Why can't you hear an aircraft traveling at mach .95 until it passes?

You can.

So to summarize, if the aircraft was going Mach 0.95, and the sound waves would slowly travel forward, wouldn't that make it so you could hear the aircraft before it got to you?

Yes, that is correct.

In a video of a F/A-18 I saw, it was traveling at what I assume was ~Mach 0.9 speeds, because you couldn't hear any sound from it until it was really close.

This observation confirms your hypothesis in the second quote, and disproves the faulty assumption in the title question.  You couldn't hear it until it was really close because it was traveling almost as fast as the sound waves it was creating.  If it were supersonic you couldn't hear it until it was actually past you.

Ah okay, thanks. But back to the cars on the highway example, wouldn't the faster car (the sound wave) get further in front of the slower one with time, meaning that you would still hear the aircraft even thought it wasn't close? Hope that makes sense haha.

Yes, but the speed of the car is much, much less that the the jet so the effect is quite different.  Actually typing this on my phone by a highway and as i watch cars approach i really don't hear them until a quarter mile or so.  Granted a jet is much louder, so at automobile speeds you'd hear it further out, but again it's way faster too ...

Now say there's a speaker or something on the nose of the aircraft facing forward. If the aircraft is traveling at Mach 0.95, it still isn't going faster than the sound waves it's making. So that means the sound waves would still slowly pull away from the aircraft going forward, right? If that's the case, wouldn't you hear an aircraft going Mach 0.95 before it passes?

Correct, you hear the airplane (or the speaker) before the airplane itself passes.
Let's suppose to be at sea level in standard atmosphere. Sound travels at 340m/s (0.2 mi/s) while the airplane travels at 340x0.95=323m/s. Let's say that you are just in front of the airplane (or the speaker) and that you begin to hear it at a distance of 3000m (1.86mi). Then you'll be hit by the sound after 3000/340=8.82s and by the airplane after 9.59s. So if you couldn't see the airplane but only hear it then you would have only 0.76s to get out of the way.

Oh okay thanks, so would the aircraft be quieter before it passes going Mach 0.95? If so, why?

Why would it be quieter?  It should make about the same sound at almost any speed below mach.

Yes, it would be quieter but for the simple reason that the sound that you hear after the 8.82s is the faint sound emitted when the airplane was 3000m away.


text:
How to pick a good airfoil based on Cl and Cd
Let's assume I have 10 different airfoils graphs of Cl x Aoa, Cd x Aoa and Cl/Cd  x Aoa. How do I pick the best one that will give me the most efficiency (the longest flight time)?
I understand the information that those graphs provide and IMHO I think I should always pick an airfoil that has the best Cl/Cd ratio at Aoa (alpha) of 0 since at 0 the drag is usually the least. This is my "common sense" but nobody says this on the internet so I am pretty sure I am wrong!
So based on the three graphs I have of each airfoil, what math should I do in order to get the most efficient (more flight time) airfoil?
PS: the plane will be a floater/slow flier/glider/sailplane. Of course it will be a slow plane cause my focus is on efficiency and fast planes usually are not efficient.

Note that the longest flight time is only ONE measure of efficiency and it happens to be the one that most designers consider not important. This is because another measure of efficiency is the furthest distance which is usually more important because most airplanes are designed to go somewhere. Interestingly, the most efficient airfoil for furthest distance is never the same as longest flight time. That's why most gliders have flaps so that you can manipulate the camber of the airfoil as a compromise

The $\alpha$=0 point on the polar would be a poor metric - there is nothing special about it, and your aircraft will not fly at this polar point when you strive for longest endurance. But your common sense is correct in predicting that this will be a slow airplane.
The precise answer normally depends on your means of propulsion. Since this is for gliders, the answer will be quite simple.
Your definition of efficiency is minimum energy loss. Energy means potential energy in this case $E_{pot} = m\cdot g\cdot h$, and the loss of potential $\frac{dh}{dt}$ energy over time is expressed as the sink speed $v_s$. So we need to find the polar point at which the glider will have the lowest possible sink speed.
Let's start with the parabolic drag equation which splits the total drag coefficient $c_D$ in one component which is constant over the lift coefficient range and one which changes with the square of the lift coefficient.
$$c_D = c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$$
 The symbols are:
$\kern{5mm} m\:\:\:\:$ aircraft mass
$\kern{5mm} g\:\:\:\:\;$ gravitational acceleration
$\kern{5mm} \rho\:\:\:\:\:$ air density
$\kern{5mm} v\:\:\:\:\:$ velocity
$\kern{5mm} v_z\:\:\;$ sink speed
$\kern{5mm} c_{D0} \:$  zero-lift drag coefficient
$\kern{5mm} c_L \:\:\:$  lift coefficient
$\kern{5mm} \pi \:\:\:\:\:$  3.14159$\dots$
$\kern{5mm} AR \:\:$  aspect ratio of the wing
$\kern{5mm} \epsilon \:\:\:\:\:\:$  the wing's Oswald factor
$\kern{5mm} S \:\:\:\:\:$  wing area
Next, we need an expression for the sink speed. This should be easy: It is the forward speed times the tangent of the glide path angle $\gamma$. If you allow to approximate the tangent function for small angles by the radian of the angle, you can write:
$$v_z = v\cdot tan\gamma ≈ v\cdot\gamma = v\cdot\frac{c_D}{c_L}$$
Now insert the drag coefficient
$$v_z = v\cdot\left(\frac{c_{D0}}{c_L} + \frac{c_L}{\pi\cdot AR\cdot\epsilon}\right)$$
and make the speed dependency of the lift coefficient obvious:
$$v_z = \frac{c_{D0}\cdot\frac{\rho\cdot v^3}{2}\cdot S}{m\cdot g} + \frac{m\cdot g}{\pi\cdot AR\cdot\epsilon\cdot\frac{\rho\cdot v}{2}\cdot S} = v^3\cdot\frac{c_{D0}\cdot\rho\cdot S}{2\cdot m\cdot g} + \frac{1}{v}\cdot\frac{2\cdot m\cdot g}{\pi\cdot AR\cdot\epsilon\cdot\rho\cdot S}$$
Now we are ready to differentiate with respect to $v$ and find the condition when the derivation will become zero.
$$\frac{dv_z}{dv} = 3\cdot v^2\cdot\frac{c_{D0}\cdot\rho\cdot S}{2\cdot m\cdot g} - \frac{1}{2\cdot v^2}\cdot\frac{2\cdot m\cdot g}{\pi\cdot AR\cdot\epsilon\cdot\rho\cdot S}$$
Now re-insert the lift coefficient, which simplifies the equation mightily:
$$\frac{dv_z}{dv} = 3\cdot\frac{c_{D0}}{c_L} - \frac{c_L}{\pi\cdot AR\cdot\epsilon}\,\overset{!}{=}\, 0$$
$$\Rightarrow 3\cdot c_{D0} = \frac{c^2_L}{\pi\cdot AR\cdot\epsilon}$$
So your energy loss will be minimized when the induced drag is three times bigger than the zero-lift drag. The polar point is then:
$$c_L = \sqrt{3\cdot c_{D0}\cdot \pi\cdot AR\cdot\epsilon}\;\;\text{and}\;\;c_D = 4\cdot c_{D0}$$
With wings of a high aspect ratio this gives a rather high lift coefficient, so you might need to pick a high-lift airfoil in order to actually trim this polar point. The figure of merit when selecting the airfoil should be the ratio $\frac{c^{³⁄₂}_L}{c_D}$; this should reach its maximum at the lift coefficient of minimum energy loss. Enter the lift and drag coefficients into a spreadsheet and create a new column for $\frac{c^{³⁄₂}_L}{c_D}$ (or $\frac{c^3_L}{c_D^2}$; doesn't matter). Select an airfoil that maximises this value at the calculated optimum lift coefficient.

I think I followed what you wrote right up to the figure of merit formula, for which I don't understand the source. Where did it come from? Thanks for taking the time to explain so much to so many.

The cl3/cd2 is the endurance parameter, a good explanation (including range related functions etc) is here : http://www.dept.aoe.vt.edu/~lutze/AOE3104/range&endurance.pdf

Page 16, (as it’s a rather long document.)

@GürkanÇetin: Yes, thank you, I could have produced the same here, but given that nobody gets the explanation for the polar point (just see what people vote on …), I found it not worth the effort.

Funny but the chosen one is a WRONG answer. The endurance is not proportional to lift-to-drag ratio, but to cl3/cd2.

@GürkanÇetin Thanks for the reference. Being self taught I blame the error on the poor quality of my instructor.

@PeterKämpf Please do not despair in helping those less knowledgeable. The answers remain for others to see. The link to your previous explanation for powered endurance is also helpful, as from other questions the subject aircraft seems to actually be a motorglider.

@Pilothead: Don't do as your instructor did but correct your answer. Others will see that it was accepted and think it is correct, learning the same falsehoods. By the way, for turbojets your answer would be correct (but not for gliders or motorgliders).

Sorry for my ignorance on choosing the wrong quesiton (I dont think it's wrong, it is just not as complete as it should). The question that I accepted was, at the time, the one that I understood better. But now that I found some time to try to understand this other answer I finally created the spreadsheet (as recommended) and I finally got the optimum point. Thanks for the great answer and I would like also to thank the other answer which first made me understand this subject.

@PeterKämpf The only way I am able to achieve the required CL for minimum sink speed is by using a flapped configuration which is reducing my endurance parameter and thus my flight time. Am I doing something wrong?

@Zaber: Please keep in mind that the answer assumes a constant zero-lift drag and parabolic induced drag. Adding flaps will increase zero-lift drag and possibly change lift distribution such that induced drag grows in non-parabolic ways as well. So the solution must be to measure the ratio of c$_L^3$/c$_D^2$ and pick the optimum of this real curve.

If you only care about efficiency, you want the one that has the highest Cl3/Cd2 for endurance and highest Cl/Cd for range at the design Cl, which may not be zero AoA. Determine desired Cl using required lift (weight plus tail download) in the following:
L = Cl * A * .5 * r * V^2
where L is lift, Cl is coefficient of lift, A is wing area, r is density and V is Velocity. Explanation is at:
https://www.grc.nasa.gov/www/k-12/airplane/lifteq.html
Look through your Cl/AoA charts at that Cl then get Cd using the associated AoA. Pick the airfoil with the highest value of Cl cubed / Cd squared for maximum endurance. Pick the highest CL/Cd using the associated AoA for maximum range.

Your answer makes sense but I didnt understand why do I have to find the L (lift). Should I first get the L or the Cl of my plane? Then using one or other I find the other right? Then I should go to the Cl/AoA (according to what you said), why? Why shouldnt I go directly to the Cl/Cd x AoA diagram?

L is the lift you want your wing to produce, the aircraft loaded weight plus tail download if that is your design. The formula gives you Cl at your design point, which you use to get AoA, which gives you Cl/Cd. You can't go directly to Cl/Cd because you don't know Cd or AoA before using the charts.

thanks for commenting. So correct me if I am wrong: FIRST I get the total weight of my plane (I will not take into account the tail download cause I have no idea how to do that and I assume the tail download will be small compared to the weight) and make L = weight. SECOND I calculate Cl (great!). THIRD: I try to find (on each airfoil) the AoA that matches the required Cl. FORTH: I go to the Cl/Cd x AoA and pick the airfoil that has the HIGHEST Cl/Cd at the chosen AoA. Is it right?

Yes. Use 10% for tail download absent anything better.

thanks! I think, using your analogy of this answer it would be possible to get an answer to the other question I did. Since with your answer I can pick the right airfoil (which I will do right now to 3 of my planes!) maybe it can be used as an iterative process to find the right chord (it will have to be an iterative process cause when I find a new chord it will change wing area and the math should be redone). Anyway my friend, you answer is by far the best one in the internet regarding this subject!

@Samul: You will not pick the best airfoil this way. Do you know the best flight speed already?

Lift to drag ratio gives the glide ratio, not the endurance performance of an air vehicle. So, although the answer has given a practical advice, it’s not the correct answer to the original question.

Flight speed is expected to be a given, as from related questions this aircraft seems to be a motorglider and no propulsion information was provided.

Answer was updated to show correct range and endurance formulas.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What Are These Runway Lights For?
I know that the strobe lights that flash in rapid succession of each other indicate the open end of the runway, to help guide pilots in at night. It's obvious the white lights indicating the edges of the runway. And I'm guessing the green lights at one end, indicate the open end, whereas red lights at the other end indicate you are approaching the end of the pavement.
But what are those lights at the ends of runways on (typically red) poles. Typically in large sets with one set on either end? Are these just to help the pilot line up? I would assume the actual runway edges and center lighting would assist with that?

As for what they are you can find that in the answer to the question here. In terms of how they work its an extension of the basic runway lighting system. There is a nice video discussing the various types here but effectively they help guide the pilot to the runway by providing them with an extended way to infer where the centerline is (beyond the runway its self) when making a low visibility approach. The system may also include a decision bar which helps a pilot by having a visual marker at a fixed distance from the threshold. Keep in mind that threshold lights lights count as runway in sight so the extended centerline can help a pilot lineup when they can not see the full length of the runway.
The colors will vary based on what the airport in question is equipped with but green is the runway threshold. Red is typically an area in front of the runway, it could be a displaced threshold or it could be an area off the runway all together. However these lights will appear on both ends of the runway and will be visible when approaching from that direction.
It should be noted that while certain lights are required for certain types of operations, lighting is not explicitly required by the FAA and there are plenty of unlit airports as well as a wide variety of lighting situations out there. The field I fly out of has REIL and 5 lights on each side of the runway which is far less than a class B airport would have but may be far more than your back country grass strip has.

It's part of the Approach light system used to help pilots align with the runway. From this FAA Page

...provides visual information to pilots on runway alignment, height perception, roll guidance, and horizontal references for Category I Precision Approaches.

How would they help with height perception and roll guidance? I don't doubt that they do, just literally HOW do they is what I'm asking?

@Jon The lights are arranged in a distinctive pattern, so pilots learn what a low/correct/high or left/correct/right or banked/level approach looks like and will reflexively fix it (make the lights "look right") before the conscious mind even realizes there's a problem.


text:
Chicago O’Hare - Two Towers?
Why does ORD have two towers? I flew in there for a connecting flight this last weekend and saw two towers standing, one obviously larger than the other.

Apparently there are four. see https://aviation.stackexchange.com/questions/27134/what-is-the-division-of-tasks-among-the-control-towers-at-chicago-ohare-airport

Simply put, a lot of airports, especially ones that are Class B (very busy), will have multiple control towers. Most of the time, there is a single operational control tower focused on departures/arrivals into the airport environment with one or multiple smaller ground control towers focused on ground movement area coordination. I hope this helps!


text:
Could I Legally Fly a Paramotor as a Commuter Method
I get permission from company security to fly into work using my paramotor (I don’t have one, but would like to get one).
We have six offices around the city. The city has two airports: one major international to the northwest of downtown and one smaller regional in the middle of downtown.
Would I be legally allowed to fly OVER the city (avoiding the downtown skyline) to make a direct route to my destination office for the day? What are regulations regarding paramotor flight in a metro area?

Paramotor as in powered parachute? Is the city covered by Class B or C or D airspace that would also require you to have a transponder/ADS-B Out and a radio? You can look at your area here https://www.skyvectors.com and see what the airspace is like.  I work at an airfield, and for a few weeks over the summer actually got a ride to where my airplane is kept and then flew to work. My car was in the shop, and it turned out to be less expensive paying for aviation fuel (and departure fees!) than renting a car for a few weeks.

Yes powered parachute, sry (I’d heard it also called paramotor). It’s covered in Class B airspace due to the Intl Airport to the northwest of downtown

this is a duplicate. Short answer is no, longer answer is in this question, which has an answer. Could I use a powered parachute to fly around New York City?

Thx! I’ll check that out

@Jon: Check the floor of the class B airspace.  If it's several thousand feet above the surface, you MIGHT be able to legally fly under it.


text:
Why Do Airlines Still Use Tractor-feed Printers?
Why do airlines still use those old dot-matrix tractor-feed printers at their gate counters? I’ve always thought that was hilarious.

This question is more related to the passenger side of aviation, which is better suited for travel.stackexchange.

My bad! Could a moderator move this to the travel SE?

Probably note directly related to aviation, maybe more to airports than airlines:
Well, dot-matrix printers are expensive, but durable and print at very low cost and finally carbon copies (two, three or even four) of forms could be created. And replacing them will require a lot of software adaptions.
I‘ve seen airports using text-user-interfaces (TUI via terminals), which are often working a lot of better (fast and efficient to use) then usual GUI applications. New stuff, is not automatically good.

IOW, if it ain't broke, why spend money to fix it?

Dot matrix printers aren't limited to a particular paper size, they can generate a continuous print out that doesn't need paginating


text:
Who Assigns Gates to Airliners?
When airliners arrive to major or medium airports, who is actually responsible for assigning gates and deciding who goes where?
Is that up to Ground? The Airline? The Airport Authorities/Management?

Related: How is the gate/stand communicated to airline pilots?

On a long-term basis the airline leases the rights to use certain gates from the airport authority. Then the airline decides which flights use which of their leased gates hour-to-hour. This information is transmitted to pilots via ACARS or a company air-to-ground radio frequency.
If there is a separate Ramp Control operated by the airline it is possible that they also have access to the gate assignments but I'm not sure about that.
Ground Control does not assign gates, but generally routes traffic to the gate the pilots say they're going to.

Nice answer. You could clarify, that this is only true for FAA-Land. In most parts of Europe for example apron control assigns the gate to the pilots.


text:
How Do Airline Pilots Park Right on the Line?
I’m sitting here at the airport waiting on my flight and I’ve always wondered: how do airline pilots park those big birds (737s and above in size. I can understand smaller regional jets where you’re closer to the ground with better line of sight) right on the line each time? Especially for 747/A380 pilots, where you guys are basically three stories off the ground!
Line of sight below them would be tricky I’d think. Is there a camera below the nose gear or a computer guidance system for ground ops? Or is it literally just practice and skill at driving the plane while taxiing on the ground to get that perfect precision while taxiing to/from the runway and to the gate?


text:
How Do Pilots Account for Weight Shift?
What I mean by this question is cases of airliners or private jets.
You have possibilities of luggage shifting mid-flight, people changing seats (occasional but it does happen), and on private jets, you have people walking around the cabin maybe to visit or in the case of larger private jets people walking all over the plane.
People will be of varying weight by age, size, and height. Would this not cause balance issues with shifts in weight around the cabin and cargo compartments? What typically happens to help account for these situations?
Is it as simple as adjusting trim? Or are there other considerations/operations to take into account?

Luggage shifting generally isn't a problem, since large luggage volumes are containerized and therefore compartmentalized, and in smaller jets like corporate aircraft, the luggage is confined to a small space.
With swept wing airplanes, fuel burn causes a natural and significant C of G shift as fuel C of G moves forward as fuel level declines in the outer extremities of the tanks, which is more aft.
If the plane has a center wing tank that is depleted first, the C of G shifts aft, then forward during the flight.  When you do weight and balance on a jet with a center tank, the fuel C of G curve on the weight and balance graph looks like a bit like a cursive letter Z in depicting this shift, and the curve has to respect the C of G limits at all points.
People moving around will have subtle effects on trim, but nothing like the fuel effects in a swept wing airplane, unless you have entire groups all moving at once. It's all about proportion. If you are a small two seat homebuilt, just leaning forward in your seat will give you a 5 or 10 knot increase in trim speed.
In any case, any C of G shift changes the airplane's trim nose up (slower) or  nose down (faster). If you are hand flying, you will have to adjust trim as you notice the plane pitching in response to the change in trim. Most of the time however, the autopilot is on, and it takes care of the trim on its own.
If you are hand flying and some passengers move forward or aft enough to make a noticeable change, you will just be instinctively blipping the trim to make the airplane do what you want.

Thanks for the great answer! I’ve always been curious about how much of an effect that has and in those cases how detrimental it is towards the operation of the aircraft. A friend of mine took me up in a small Piper once and mentioned how even the smallest shift in weight or C of G could cause adjustments to have to be made so it made me curious about bigger private jets and even airliners.

You rarely hand fly a jet in cruise. Just too much work. So the AP masks the changes 98% of the time.  My father flew C-47s during WW2 in Burma, doing combat support airdrops into jungle clearings near Japanese lines and such, and they would take a crew of 3 or 4 extra "kickers" to help move and push the bundles of ammo and food out the door during the drop. On the way back with an empty airplane, they would have fun by running back and forth in the cabin to make him have to constantly retrim. Bunch of 19-21 year olds, what do you expect?


text:
How does a turbojet power its own compressors?
The title says it all, does the compressor use electricity? if so, how does it generate this electricity? The main power source can't be from the compressor itself as it would lose more power than it would gain.
Not a duplicate of:
How much power is drawn by a turbofan to power its own compressor?

This question reveals insufficient prior research. Try searching for diagrams and illustrations of turbojet engine construction.

No. They don't . There is a turbine driving it.

You seem to be confusing power with electricity.

@A.I.Breveleri I have done my research, all I've found are diagrams of how the air flows, I needed to know what makes the fan actually turn.

Surely you must have noticed that the compressor and the turbine are on the same shaft. - Although, I must admit, that shaft is not explicitly labeled in any diagram I've seen. - It runs right through the middle of the combustion chamber section of the engine.

a gas turbine engine does not use electricity to spin its compressor stage. the compressor discs are driven by a geared shaft which is rotated by the exhaust turbine.
the thermodynamic cycle which forms the basis of operation of a gas turbine engine is called the brayton cycle. an analysis of the brayton cycle shows that the majority of its (gross) power output is used to operate the compressor stage. The remainder represents the useful (thrust) output of the engine.

If that's so, a compressor, like the name states, compresses air into the combustion chamber which requires a decent amount of torque, so are you saying that the thrust generates the compression as well?

@talkingmango, no: the combustion of fuel in the engine provides the expansion of the airflow through the engine; when this airflow goes through the exhaust turbine, it furnishes the torque that is fed to the inlet fan (compressor stage). what power remains after powering the compressor is used for thrust.


text:
Can one "create" a private interpilot air-to-air frequency to chat?
I've been looking for an answer to this for the past couple of hours and can't find anything on the matter. (Yes, I've read this What radio frequencies are used for intercom in formation flying?  , but it didn't answer my question, as I'd like to know about frequencies that are not in use by any authorities or  organisations and are not used for anything official,  such as search and rescue).
A pilot mentioned to me she "created" frequencies to talk in private with her mate (air-to-air interpilot). I found one link that says pilots sometimes use unallocated frequencies to chat , so I assumed she meant that, but I've just read on AOPA that all 760 channels available in the VHF   aviation band are actually allocated (I'm assuming in the USA, as the reference is an FAA circular) even if some are not used at all times.
So, can you "create" a private frequency? If I'm right to assume that that makes no sense, can you join a private frequency (do they exist??) or use an unallocated one to chat privately? In case it's the latter, how does that work if all frequencies are allocated already (or is that just in the USA)?
Thanks a lot for your help

There are frequencies set aside for air to air conversations but they are not private.  Any other person with a radio tuned to that freq can listen or talk.

Private assumes that the conversations are not comprehensible for anyone else on that channel. This would require encryption on both ends. If you can manage that, then it doesn't matter who else is listening on the frequency. Is this what you are aiming for?

The best you can practically get is to use (share with everyone) a unlicensed frequency, and encode/encrypt it in a robust way so that sharing is not a problem, just like how WIFI and bluetooth does it.

I don't think this answers your question, but 122.75 is designated as the fixed wing air-to-air frequency and 123.025 for helicopters. While they are to be used for the purpose you outline, though without any suggestion of privacy.

When I was earning my licenses in 1975, the FAA provided each of the flight schools (4-5) at the airport I flew out of, a discreet frequency and all training aircraft where required to monitor their FBO base station frequency once they were out of the pattern. I assume this provision is still available in some form.

No.
Within the United States the designated air-to-air frequency intended for this purpose is 122.75 MHz. If you use a different frequency you could interfere with some other aviation band communications somewhere else within range of your transceiver. Do not do this.
See the Aeronautical Information Manual (AIM) 4-1-11 in table 4-1-3 for other related frequencies but 122.75 is the standard go-to.

While this is indeed aviation-related and the FAA publishes documents listing frequencies, allocation and reservation of radio frequencies is handled by the FCC, not the FAA. It is noteworthy that while some frequencies are agreed upon by international treaty (read: ICAO), each member country’s government is responsible for frequency allocation within that country. In the U.S. that’s the FCC.
Source: I’m a flight instructor.

The FCC allocates the overall band but the FAA has a department just for managing specific station frequencies within air band. The FAA department is "Spectrum Engineering AJW-1C" and they handle both navaid volumes and comms allocation along with the help of the flight inspection department.(flight check maps the actual signals in use)  The FCC and FBI assist AJW-1C when there are issues that need investigation like artificial interference.

Every frequency is allocated somewhere, in fact the same frequency will be used over and over throughout the world. This works because air band radios are relatively short range, about 200nm at best, so all you really need to do is find a frequency that isn't used anywhere near you and tune to that. This is what formation or group flights do.
That will not be private in the sense that it cannot be overheard, anyone with an air band radio can listen in. It will also not be private in that only you could use it, again anyone could tune into the same frequency for the same purpose. Formations often have secondary frequencies to tune to in these cases, but it's rare they need them.

123.45 is commonly used as an (unallocated) air-to-air frequency, although depending in your location it might not be officially sanctioned
You may use that for chat or formation flying, but it will definitelly not be private

Within the United States, 123.45 MHz is authorized to be used only non-governmental flight test operations, and is not to be used as an air-to air frequency. However, over remote and oceanic areas, out of range of ground stations, it is designated as an air-to-air frequency. See FAA Order 6050.32. See also the FCC regulations for more information on frequency designations and penalties for misuse. Note that 122.75 is designated as the fixed wing air-to-air frequency.

Huh... That sounds like it should have been an answer...


text:
How bad is the efficiency of jet engines under low loads?
VTOL operations need 4-5 times more thrust than a conventional take-off of an aircraft.
If you don't want to install additional lift-engines into a hypothetical aircraft and use thrust vectoring to turn the thrust towards the ground, you naturally end up with largely oversized engines for the cruise flight.
But how bad does the fuel efficiency (thrust per amount of burned fuel) get, when the engines run at their minimal self-sustaining throttle setting during cruise?

A thrust/weight ratio around 1 is standard for fighters, so making them VTOL does not result in any larger engines than otherwise!

Sorry, I think my question was not precise enough. IMO VTOL operations are not only beneficial for fighters, but also for small to medium sized civil aircraft, such as an equivalent to a rescue helicopter. Normal airliners or private jets usually have thrust to weight ratio of 0.3-0.4.

In this specific case, there no high thrust needed to sustain a speed well below the speed of sound and without any spectacular maneuvers.

Efficiency of a jet engine is pretty bad at low throttle settings. Idle rpm is something like 70% of max. rpm.

A VTOL aircraft needs a 1:1 T/W ratio at maximum takeoff weight. Fighters don't usually achieve that, they only get to 1/1 T/W at typical air-air combat loads.

Current jet engines are actually not built for efficiency rather than performance.
If you consider efficiency, you don't build a VTOL aircraft. This is why the F35 has multiple versions with and without VTOL capabilities.
Also, jet engines even during flight have to be able to sustain fast speed change, fast altitude increase and high load factors, all combined with lightweight designs. Therefore, I wouldn't consider takeoff as the most demanding phase of flight for such engines, but I may be mistaken on that point.
All these conditions make it difficult to reach the efficiency of say turbofan or turboprop engines.

Note that the term “jet” engine usually includes turbofans. And nobody's making pure turbojets anymore anyway—fighters usually use turbofans with low bypass ratios 1:0.5 to 1:1 these days.

@JanHudec, totally agreed

@JanHudec the opposition between jet egines and turbofans was more about their usage. No high bypass ratio turbofan would sustain the movements of a fighter aircraft for long

There is no opposition between jet engines and turbofans, because turbofans are jet engines. They are not turbojet engines, but they are jet engines as most people understand that term.

@JanHudec what do you mean by they are not turbojet engines ?

A turbojet engine is one where all air flow goes through the hot section. A turbofan is an engine that has a bypass, cold, flow. A jet engine is a vague term usually considered to cover both types. So a turbofan is not a turbojet, but it is a jet.

Extensive research was done into civilian VTOL operations in the 1960s and 1970s. Those projects were killed by several factors:

high fuel cost (combined with the fuel crisis in the early 1970s)
high capital cost (due to all the extra lift engines needed compared to regular aircraft, there were designs with dozens of lift engines).
noise. The attraction of VTOL is being able to operate near a city center instad of having to drive to a remote airfield with a big runway. But all that thrust necessary for VTOL results in an awesome racket (the Harrier is one of the loudest fighers ever, despite having much lower performance than its contemporaries).  Nobody is willing to have that much noise in an urban setting, making the whole exercise pointless.

The main problem with installing bigger engines is that they are heavier. And because you need to lift the engine weight too, the thrust/weight ratio grows slower than the installed thrust. The result significantly reduced payload fraction and payload fraction is the major factor in efficiency.
Now fighters normally do have thrust/weight around 1, but they are basically aerodynamic shells wrapped around the engines and some fuel tanks and almost nothing else. Their payload is too small compared to transport aircraft to be practical.
Now vertical take-off is occasionally useful in civilian operations, but with all the airports already built not as often. For the cases where it is we have the rotorcraft and the AugustaWestland AW609 tiltrotor.
Unlike jets, the large rotors have much lower induced power, so they can produce much more static thrust with the same power and thus support hover with reasonably small engines. The price is that their thrust decreases faster with speed, so the top speed is lower.

https://en.wikipedia.org/wiki/Harrier_Jump_Jet
https://en.wikipedia.org/wiki/McDonnell_Douglas_F-15_Eagle
Both are twin engine fighters and have similar thrust - one appears to be a litte more aerodynamic than the other.  So I would think that fuel efficiency when
minimal self-sustaining throttle setting during cruise is used could be similar too.

The Harrier is a single engine aircraft

Ok, comparison can be made to the F16 instead.


text:
Why does the centrifugal twisting force move variable pitch propellers to fine?
In my research on the forces acting on and influencing a single engine, non feathering variable pitch propeller's ability to change blade angle, I've come across a bit of a roadblock.
Aerodynamic forces try to move the blades into a coarse pitch, as the blade's centre of pressure is ahead of the blade's pivot point. That much is obvious and makes perfect sense.
What stumps me is the centrifugal twisting force. How does this send the propeller into a fine pitch?
Several sites talk about this concept, but only to the extent that they acknowledge it exists, and in what context it occurs. Here are some examples of the only relevant sections  per site:

Centrifugal twisting force, being greater than the aerodynamic twisting force, tends to force the blades toward a low blade angle.

Source

The centrifugal force acting on the blades tends to twist the blades such that the blade angle reduces. This force is somewhat opposed by the aerodynamic twisting force.

Source (Also comes with this image that has no explanation)

Centrifugal And Aerodynamic Twisting. Any asymmetrical spinning object generates a centrifugal twisting force, the propeller is no different with the force of its spinning action twisting the blades to a fine pitch.

Source
I could go on.
Given that the force refers to 'centrifugal', I suspect centre of mass has something to do with it. If so, where is the centre of mass located? Forward, or aft of the pivot point? Seems like it would be slightly forward, given that a blade's airfoil shape would require more mass at the leading edge.

Reference

"centrifugal twisting force" cannot be understood if you don't provide the principle used for pitch adjustment. The centrifugal force acts on some mass, which in turn can twist the blade. To understand how this happens, you need to look at and disclose here the mechanism used.

The centrifugal twisting force in variable pitch propellers does not come from the blades themselves. It is induced by ballasts connected to the root of the blades via a mechanism such that when the spinning motion of the propeller induces a force on the ballast, it will (in most cases) force the blades into fine pitch.

Do all propellers have this? I've seen some with fairly obvious counterweights, but nothing on others. Unless they are hidden behind the spinner? The confusion I have is that if the centre of mass is behind the point where the blades pivot for pitch, it would make sense, but without those counterweights, CoM would have to be ahead of the pivot and would want to move coarse instead, as the blades would be thicker closer to their leading edges.

No, on small aircraft adjustable propellers are not very common, as they are quite expensive, bot to buy, and to maintain. For "hobby" use they offer little advantage for the price, but for passenger planes they add considerable amount of economy and safety in form of much improved efficiency and performance.

If you mean do all adjustable propellers have the counterweights, I mus say I'm not certain. The weights can be quite small however, and may be hidden inside the spinner.

There (indeed) are single acting and double acting blade fineness adjustment systems.  In the single acting design, the hub counterweight is set to use centrifugal force to move blade to coarse.  Oil pressure, in the form of a hydraulic piston, is used to move the blade to fine.
In a double acting system, oil pistons move the blade both ways.
Confusion may arise in the statement "moves the blade to a lower angle", as for feathering: the "low blade angle" is to the wind, not to the axis of rotation (very coarse).
For the propeller blade itself, it is entirely possible that more mass may be ahead of the pivot point which would cause a centrifugal twist force towards fine.

"Aerodynamic forces try to move the blade to a coarse pitch"

Yes, and this is counterbalanced by the blade center of gravity trying to move to fine from centrifugal force.  This helps balance forces on the blade shaft.
But we must consider all components of the rotating plane, including the hub and attachments.  In order to be able to "feather" the prop in the event of engine failure, a counterweight is added to the hub to fully coarsen the prop pitch.


text:
What is the main cause of pressure recovery on airfoil?
What is the main cause of pressure recovery over a wing? I’ve narrowed it down to be one of or a combination of 2 things:

The higher pressure air from the bottom of the wing equalizing with the upper surface pressure, explained here clearly.
The atmospheric air pressure equalizing with the air pressure on top of the wing.

So is it both of these that cause pressure recovery? Does one contribute more than the other?
(If I had to guess, it probably depends on the scenario, high AoA, maneuvering, etc)
(It’s also possible there’s another cause I missed and don’t know about)

I don't love your wording, but it is essentially your bullet two.  But your bullet one is also true.
Although the pressure can increase / decrease when the flow accelerates around the body, it must eventually return to the ambient pressure.  At the trailing edge, the top/bottom pressure must be equal.
For subsonic airfoils, you essentially reach ambient pressure  at the trailing edge.  You can be a bit off, but you're pretty close at the TE.

I see. So for supersonic airfoils the pressure isn’t at ambient when you get to the TE? (I know for subsonic airfoils it isn’t either, but subsonic airfoils are closer to ambient)

The TE shocks on a supersonic airfoil satisfy the condition that the upper and lower pressures are equal and the TE velocities are parallel.


text:
Why will a slower moving parcel of air above a wing create drag in non-inviscid flow?
This answer explains why a thicker boundary layer will cause a type of form drag. It uses inviscid flow to explain this.
It is a great answer, but I can't seem to see how this applies in a real world scenario. Because flow is not inviscid in actuality, how does this work? (Ex. real flow has viscosity and drag, the BL doesn't extend forever)
I'm not very experienced with inviscid flow, so that answer is a little hard to understand, but this is what I've gathered so far:

The slower moving BL on top of the wing will create less skin friction drag than if it was moving faster. Pretty simple.
If a faster moving parcel of air 'speeds up' the boundary layer, then that would make more drag, because it causes more skin friction drag.

I'm not sure if this is what the linked answer was essentially saying, but it's what I've understood.
The linked answer also says:

From all this, we see that form drag is a pressure drag -- but it is a pressure drag that only exists because of viscous effects (the presence of the boundary layer).

Which also helped me come to that conclusion. (Not sure if it's the right conclusion or not)
To summarize, real flow isn't inviscid, so how does this work in a real situation?

My answer above may have missed your question a bit (I found your question hard to identify...).  Here is another answer that may help.
Imagine a super-thin flat plate airfoil at zero angle of attack.  The top and bottom surfaces are in contact with the flow.  We call these surfaces 'wetted'.  The wetted area is twice the chord times the span.
If the flow were inviscid, this thin flat plate at zero alpha would be 'invisible' to the flow.  It would not cause any accelerations.  It would not cause any turning.  It would just sit there having no effect.
If we measure the drag on this two-sided (top and bottom) thin flat plate at zero angle of attack, we get pure skin friction drag.  Sometimes we even call this the 'wetted area drag'.
If we could measure pressures along the plate, we would find that pressure is equal to the freestream static pressure everywhere along the plate ($P=P_\infty$) -- pressure coefficient is zero everywhere along the plate.
Recall:
$$C_p=\frac{P-P_\infty}{0.5\,\rho_\infty\,{V_\infty}^2}=1-\left(\frac{v}{V_\infty}\right)^2$$
Next, we 'inflate' the flat plate to make it into a reasonable symmetrical airfoil shape.  It is still at zero alpha (zero lift).  It has the same wetted area as our flat plate.
Now, as we know, the airfoil causes acceleration in the flow.  It causes a pressure distribution.  At most locations around the airfoil, the pressure coefficient will be non-zero.
The drag on the thick airfoil (non-lifting, still at zero alpha) will be greater than the drag on the flat plate (of the same wetted area).
This difference is the form drag.
If you make the airfoil thicker, the flow will accelerate more, the form drag will increase.  If the airfoil is thinner, the flow will accelerate less around it, and the form drag will be less.
The non-lifting flat plate only has skin friction drag -- also called wetted area drag.  We call it wetted area drag because we can calculate it by calculating the wetted area and the Reynolds number.
Some might say that the thin flat plate airfoil has no shape to it.  It has no 'form'.
When we thicken the airfoil, we give it shape.  We give it 'form'.  The increase in drag (over the drag of a flat plate with equal area) is the form drag.  The form drag is caused by the super-velocities (as compared to a flat plate) around the airfoil caused by its thickness.

Bottom line...
The pressure distribution around an inviscid closed 2d curve (airfoil) is such that all the pressures cancel and the pressure drag is zero.
As you point out, that is not the real world.
When you consider a real viscous flow, the pressure distribution is changed and they no longer cancel and the pressure drag is non-zero.
All the rest about control volumes and boundary layer thickness and inviscid layers on the outside of boundary layers, etc.  All of those things are mathematical abstractions and approximations that we use so we can analyze and calculate the flow using simplified tools.
In the early 1900's aerodynamicists developed powerful techniques for analyzing 2d inviscid flows.  They were also developing techniques for understanding boundary layers.
These techniques are powerful, but they are also very much simplifications (inviscid incompressible flow for example).
In reality, all of the different kinds of drag (form, induced, wave, trim, cooling, interference, etc).  Are all just mathematical abstractions that are related to the mathematical tools we've developed to analyze these systems.  In the real world, all aerodynamic forces are either pressure or shear.  That is it.  Pressure or shear.


text:
Why will the pressure distribution of a swept wing promote stall?
Here it explains how the bound vortex will change the lift distribution of a wing. What is meant by the lift distribution changing is that there is less lift at the tip, meaning the pressure differential between the 2 surfaces is less towards the tip. I think I understand this part.
However, I do not understand how this will make the tip stall first. If the pressure is higher (higher compared to the wing root upper surface pressure) on the upper surface toward the tip, I always thought that would make a stall happen easier.
For example, a lower pressure on the upper surface would cause the pressure recovery to not be handled as well, encouraging stall.

So to summarize, the bound vortex through some things explained in the linked answer will change the pressure distribution on the wing. It does this in a way where the pressure differential is less and less as you move towards the tip. Why would this pressure distribution promote stall?

This answer will help.

Are some of the usages here of stall an abbreviation for tip stall, or are they really "stall"?

The lift distribution is given by:
$$l(\eta)=c_l(\eta)\,c(\eta)\,q$$
Where $\eta$ is the spanwise coordinate.
The notation $l(\eta)$ is meant to emphasize that the local lift $l$ is a function of $\eta$ -- i.e. it varies with $\eta$.  Likewise, $c_l(\eta)$ and $c(\eta)$ are meant to emphasize that the local 2D lift coefficient $c_l$ and the local chord $c$ both can vary with span.
Often, we will divide by the dynamic pressure.
$$\frac{l(\eta)}{q}=c_l(\eta)\,c(\eta)$$
Aerodynamicists will use the quantity $c_l(\eta)\,c(\eta)$ as a scaled version of the lift distribution.  Sometimes, they divide by a reference chord $\bar{c}$
$$\frac{l(\eta)}{q\,\bar{c}}=\frac{c_l(\eta)\,c(\eta)}{\bar{c}}$$
This still gives us a curve with the shape of the local lift distribution, but divided by two constants such that it is scaled to be similar to aircraft lift coefficients.
The important part here is that a non-uniform lift distribution $l(\eta)$ (which is all of them) means that the local lift coefficient $c_l(\eta)$ and or the local chord $c(\eta)$ also have a non-uniform distribution.
In fact, if the lift is large at the tips, that means that the local lift coefficient and/or the chord must be large at the tips.
Or, if the wing tapers (chord gets small at the tips), then even a uniform lift distribution will cause the lift coefficient to increase at the tips.  We can see this by one more algebraic manipulation.
$$\frac{l(\eta)}{q\,c(\eta)}=c_l(\eta)$$
Here we see that the local lift coefficient depends on the ratio of the local lift and the local chord.
OK, what does all this mean?
Here is a swept wing (30 deg at the LE) with some taper ($\lambda$=0.5):

Here is the resulting lift distribution:

The lift distribution (cl*c/cref) is in yellow, and the lift coefficient (cl) is in red.
The local lift coefficient represents how 'hard' that part of the wing is working.  Let's assume that this particular airfoil is expected to stall at $c_l$=1.0.  That is a 2D number -- and clearly this is a 3D flow.  However, we can use the sectional lift coefficient compared to the 2D stall lift coefficient for the airfoil as a simple measure of how close to stall you are.
We see that this wing (at this angle of attack) reaches a peak sectional lift coefficient of about 0.887 at a span location of Y=2.8.
In terms of your question, the peak sectional lift coefficient is relatively outboard -- which means the wing will likely stall at that section (near the tip) first.
Here is the same chart for an un-swept wing (zero degrees at c/4).

Notice how the location of the peak lift coefficient moved inboard?  Also notice how the lift coefficient distribution is much flatter?
Here is the un-swept wing, with the taper eliminated.  Consequently, the yellow and red curves are identical.:

For an un-swept and un-tapered (and un-twisted) wing, the peak of the lift distribution is at the center line.  This will stall from the center out, producing almost no rolling moment from the stall.
And, for completeness, here is the swept, but constant chord wing, again the red and yellow curves are identical.

As you can see, sweeping the wing changes the lift distribution and moves the peak of the lift coefficient outboard.
Swept wings are usually combined with tapered wings, which further pushes the peak of the lift coefficient outboard.
These things can result in wings with poor tip stall behavior.
Edit: Include image of line vortex influence.
When we talk about potential flow, we talk about how certain flow structures (sources, sinks, vortices, doublets, freestream) cause a velocity field around them.  We combine many such flow structures to assemble a complex flow.
In the math abstraction, we say that the vortex causes a velocity field around it.  However, when we talk about a wingtip vortex, we don't really believe the causality works the same way.  The flow results in the wingtip vortex, not the other way around.  I know this is confusing.
Here is an image depicting what a line vortex does.  A vortex of a given strength causes a velocity field around it.  Notice that a vortex has no influence on itself.
So, the trailing vortices cause a velocity increment at other places on the wing.  So their influence is felt someplace else -- not at the location of the vortex.

I see, thanks a lot for your detailed answer. So as for a swept constant-chord wing, why will the maximum lift be made not at the center of the wing, or at location -2 / 2 in the last graph? If the chord is constant, why is this the case? (Not considering spanwise flow, unless that is the reason)

Recall your question that spawned this one.  Sweeping the wing moves the horseshoe vortices aft and changes the lift distribution.  The center of the wing is least affected by the trailing vortices.

Right, but in the last graph, the horseshoe vortices seem to be affecting -2 / 2 most, not the center where the horseshoe vortices should have the most effect, correct? My theory is that this is due to spanwise flow.

it would seem that the trailing vortices are causing an upwash near +-2 that causes the local alpha to be higher and therefore higher local lift.  These vortices have less effect near the center line.

I see. I'm assuming the reason for less effect near the centerline is due to the sweep reversing and changing the flow pattern? Sweep reversing meaning where the wing sweeps the other way at the centerline.

Don't you recall the other question where we talked about how the trailing vortices mostly have an effect perpendicular to them?  Therefore, there are many such vortices affecting the outboard half of the wing, and fewer affecting the inboard half?

I do recall that answer, but it said "When you apply these vortices to a swept wing, the inboard trailing vortices start first. They are in-line with the wingtips and have greater influence on the lift of the wing. The vortices that trail from the wing tips start later and do not have the same alignment with the center of the wing in order to have an effect." which I thought meant the inboard vortices have more of an effect than the outboard. I think there must be something I'm not understanding here.

(I will admit I don't understand bound vortices / horseshoe vortex systems that well)

I just posted an edit and image that might help.


text:
Why will the max lift point not be at the center of a swept but constant chord wing?
(In this question when I say "swept wing" it means the blue wing seen at the bottom. So no fuselage or anything) If you look at this graph from this answer, you'll see the max lift point of this wing is not in the center. This wing is a swept but not tapered wing:
I studied this answer and also this one for an afternoon and I can't seem to understand this. I think this is because I don't really understand the horseshoe vortex system / the bound vortex.
Consequently, I don't understand how a swept wing with no taper will have its lift fairly dramatically drop off right in the center of the wing, while the whole wing is essentially symmetrical. (This isn't considering spanwise flow)
It's hard to see how the bound vortex system translates to actual flow over a wing, if that makes sense. So the question is, why does this lift dip in the center of the swept but non tapered wing happen?

Unless it's because of this: An airfoil will create low pressure. Higher pressure freestream air will tend to be attracted to this air. As explained in the 3rd paragraph of this answer. Knowing that, air in front of a wing will bend towards the leading edge. If the wing is at an angle, this air will bend to where it's in line (mostly) with the LE, as seen in the picture below.

Because the left side of the swept wing will bend the air inwards, and so will the right side, the air on both sides of the swept wing will intersect where the red lines intersect. When this happens, it could make some turbulence or interference that could cause the dip of the graph seen in the first picture

This is the only reason I can seem to think of, although it might not be correct.

This question is in a way asking if there is a way to explain this without using bound vortices, as I’m having some trouble understanding that.

It's the Mitteneffekt.
Remember that air gets sucked into the region with lowest pressure, so the streamlines get bent towards the center on the upper surface of a swept wing. This will hit an obstacle near the center: Here, streamlines converge and "fill up" the low pressure area, resulting in a higher pressure at the wing center. This explains the rise in the pressure coefficient at the center of the wing in your first plot.
If you now look at the distribution of horseshoe vortices, their intensity is proportional to the local lift coefficient. A drop in the middle is caused by pairs of horseshoe vortices which extend only over parts of the inner wing, with the free vortices trailing off not near the tips, but one near the center and the other at the outer wing. In a very crude plot this looks like this:

Horseshoe vortices on a swept wing, own work. Note the direction of rotation: The innermost free vortices indicate a drop in circulation strength at the center of the wing.


text:
Why do military and tandem seat aircraft have the throttle on the left?
Why do military and tandem seat planes have the throttle on the left? As seen in the photos below of an F18 Hornet and a P38 Lightning. Is there a reason for this layout?

It's got to be one way or the other.  Why would you not want right hand on primary control (stick / yoke ).

@quietflyer Most of us fly from the left seat and control the yoke with our left hand and radios, prop, throttle from with our right. I’ve flown from the right seat a few times and it feels unnatural.

I couldn't find any documentation, but the most logical answer is most people are right hand dominant, and prefer the primary airplane controls in that hand where the placement is optional, so stick right, throttle and other secondary controls left.
Why side by side seating didn't put the PIC in the right seat like most helicopters? That's a bit of a mystery.  Plane and Pilot magazine  says side by side a/c put the PIC in the left side because of the need to have a better view of the runway left hand traffic patterns.
Another one is oncoming aircraft are supposed to yield to the right, so it makes sense to sit on the left to better observe oncoming a/c.
If sitting on the centerline of the airplane where none of that matters however, most pilots are going to prefer the stick in the right hand with the stronger arm muscles.

"Why side by side seating didn't put the PIC in the right seat like most helicopters?" - maybe was inherited from cars, without anyone really thinking it through?

Potential ASE question: what were first (numerically significant) instances of side by side seating w/ dual controls, with shared throttle in middle, and was primary pilot on same side as driver sat in automobiles of that nation?

I think car conventions might be part of it.  Helicopters are interesting.  You really need to have the stick in the right hand whichever side you sit, so turbine Bells and some others fly from the right so you don't have to switch hands to work switches and radio.   Others  you fly from the left and you have to switch hands.

The reason may have to do with a desire for the pilot to maintain constant control of the aircraft with the right hand while allowing the left hand to perform cockpit tasks while not resting on the throttle.  This configuration is also adopted by helicopters with the pilots right hand controlling the cyclic and the left hand controlling the collective.  There is no reason that the reverse of that cannot be accomplished i.e. controlling the aircraft at all times using your left hand and performing cockpit tasks with your right.
Strangely enough, the tanker position lights on the bottom of aerial refueling tankers were originally set up your mind with the requirements of a captain sitting left seat.  The lights on the left side represent positioning up or down relative to the tanker, and the lights on the right side represent positioning forward or aft of the tanker.  This was done primarily because SAC bomber pilots operated large bombers with their left hand on the yoke and right hand on the throttles.  So it made sense that if you wanted the guy to climb relative to the tanker you would mark the low position light on the left side back to have a pilot instinctively pull back on the yoke to rise the aircraft to align with the tanker correctly, etc.  Experiences in Vietnam showed that the use of refueling tankers and tactical aircraft had been somewhat of an afterthought in cold war strategy.


text:
Who and where can access the list of ICAO24 addresses?
Who and where can access the list of ICAO24 addresses?

The blocks of ICAO addresses assigned to each state can be found in this document. How each state assigns its block to individual aircraft is up to the state, and there is no authoritative source where you can look up any worldwide ICAO address.
The United States, for example, uses a simple map of tail number to ICAO address. They are assigned all addresses from A00000 to AFFFFF. A00001 is given to N1, A00002 to N1A, A00003 to N1AA, and so on in order up to ADF7C7 for N99999.
It may be puzzling that no such map exists for aircraft worldwide. But from an operational standpoint it doesn't matter which address is mapped to which aircraft. The only thing that matters is that no two aircraft share an address.


text:
Weight limits for airliner bogie 6 landing gear
I have been unable to find the weight limits for 6-wheel bogie landing gear? I did find that for 4-wheel bogie landing gear, is for aircraft 200,000 lbs or greater. But what is the limits for 6-wheel bogie landing gear? Thankyou for the help.

I haven't looked into the specific limits, but as it's the load per wheel that is limiting the weight, a first guess would be 6/4 * 200,000 lbs


text:
Is there a public database for registration of jet engine ownership?
I am looking for a way to track prior ownership of a jet engine to confirm I am buying from the holder of record.  Does such a database exist?

The original engine manufacturer may have some records.  Whether or not they are willing to share them with you, I don't know.  I am not aware of any public database.

Does such a database - i.e. a public database - exist for buying, say, a used car? As hard as it is to "prove a negative", it may be difficult to get a supported, definitive answer to this question, but I'd strongly suspect that the answer will end up being "no".  A local FSDO might have the expertise to be definitive about this.

@RalphJ in some countries, such things do exist for cars. They won't list who owned it, but will list things like accident history and clocked distance traveled so you can check if the paperwork is correct and the odometer hasn't been tampered with.

I visited a salvage yard for engines near Harford CT years ago. They had a large number of J-47 ( ? ). Amoco used them for stationary power. There was little apparent record keeping.  The deal was they "guaranteed" the engines , If you bought one and it failed , they exchanged it for another .

This information is not kept in a centralized public database. Manufacturers will also be unlikely to be able to answer such a query because they don't normally keep track of the transactions in which the ownership of the engine changes hands, except maybe indirectly if the engine has been under a continuous maintenance contract with them. I also doubt they would make this information publicly available even if they had it. This data is supposed to be kept for each engine in the engine log.
You may be able to indirectly determine that an engine has been with the aircraft for some time by requesting a CD from the FAA with the records of the aircraft in which it is installed, but that would also be a longshot.
If you are just buying an engine as a spare or a replacement, you should closely examine the engine log anyway. If chain of ownership is not clear from the log entries, or if there are unexplained gaps in the logs, then you can decide if you want to proceed with the transaction.


text:
How does the position of Maximum camber affect lift and drag on an airfoil?
I am trying to understand how varying the maximum camber position of an airfoil will affect the lift and drag it generates. For instance considering NACA 2415 having maximum camber at 40% of cord length how would it be different to having maximum camber at say 47% cord length?

It mostly affects the pitching moment.

Do you really mean maximum camber? The thickest part of the wing is not necessarily the point of maximum camber. Many airfoils have constant camber over much or all of their surface, and only the thickness, and hence curvature of the upper surface, vary substantially.

Well, move the camber back by dropping some flap.  You will see more lift, but also more drag.  Lift center of pressure of the wing may also move back, but downwash on the tail may also increase. Placing maximum camber at 40% chord (while placing maximum thickness at 30% chord) is accomplished by hollowing out the bottom of the wing a bit underneath towards the trailing edge.
The effect of this would be to improve the Coefficient of Lift by increasing "bottom" lift while minimizing drag increase as compared with a fully symmetrical airfoil.  Years of work in wind tunnels seems to yielded the "30/40" as a very good lifter and cruising wing for aircraft in the low subsonic range (up to around 250 knots).  The Davis airfoil is one of the more famous examples of this design.

What a simple, practical demonstration!